Suppose two forces represented by the vectors $F_{1}=\langle -2,1,0 \rangle$ and $F_{2}=\langle 1,1,-2\rangle$ are acting on a particle in $\mathbb{R}^{3}$.
If the particle were to remain still, I found that the force vector $F_{3}=\langle 1,-2,2\rangle$ would need to act on the particle in combination with the other two.
Now, suppose instead that the third force vector is $3F_{3}$ and that the particle moves in a straight line from $(1,0,-2)$ to $(0, 2, 1)$. I need to find how much work is done by the combined forces $F_{1}$, $F_{2}$, and $3F_{3}$.
First of all, I'm confused about what they mean by $3F_{3}$ here. Is it $3$ times the same vector $F_{3}$ that I found that kept the particle from moving? Is it the vector from $(1,0,-2)$ to $(0,2,1)$ with components $\langle -1,2,3 \rangle$? Or is it a new vector I'm supposed to solve for by adding up $F_{1}+F_{2} +3F_{3}=\langle -1,2,3\rangle$? Or is it something else entirely?
Then, once I have all of this information, how do I find how much work is being done? I'm sure there's a formula, but I don't have a book, and whenever I Google "work formula", stuff that has to do with integral a keeps coming up which isn't terribly helpful. So, I'm asking you to do me a solid and just let never know what the formula I mean supposed to I use is.
Thank you ahead of time.
$3F_3$ means $3\langle 1,-2,2\rangle = \langle 3,-6,6\rangle$. But they want you to figure out the work done by the force $$F_1 + F_2 + 3F_3 = \langle 2,-4,4\rangle$$ If you're not taking a calculus-based physics course then they must define work as $W=F\cdot d$ where $d$ is the displacement vector. If you are taking a calculus-based physics course then you'll want to know the actual definition (which does involve an integral) even though in this particular problem it'll still reduce to $F\cdot d$.